Problem: Simplify the following expression: $y = \dfrac{-9x^2- 29x- 6}{-9x - 2}$
Answer: First use factoring by grouping to factor the expression in the numerator. This expression is in the form ${A}x^2 + {B}x + {C}$ First, find two values, $a$ and $b$ , so: $ \begin{eqnarray} {ab} &=& {A}{C} \\ {a} + {b} &=& {B} \end{eqnarray} $ In this case: $ \begin{eqnarray} {ab} &=& {(-9)}{(-6)} &=& 54 \\ {a} + {b} &=& &=& {-29} \end{eqnarray} $ In order to find ${a}$ and ${b}$ , list out the factors of $54$ and add them together. The factors that add up to ${-29}$ will be your ${a}$ and ${b}$ When ${a}$ is ${-2}$ and ${b}$ is ${-27}$ $ \begin{eqnarray} {ab} &=& ({-2})({-27}) &=& 54 \\ {a} + {b} &=& {-2} + {-27} &=& -29 \end{eqnarray} $ Next, rewrite the expression as $({A}x^2 + {a}x) + ({b}x + {C})$ $ ({-9}x^2 {-2}x) + ({-27}x {-6}) $ Factor out the common factors: $ x(-9x - 2) + 3(-9x - 2)$ Now factor out $(-9x - 2)$ $ (-9x - 2)(x + 3)$ The original expression can therefore be written: $ \dfrac{(-9x - 2)(x + 3)}{-9x - 2}$ We are dividing by $-9x - 2$ , so $-9x - 2 \neq 0$ Therefore, $x \neq -\frac{2}{9}$ This leaves us with $x + 3; x \neq -\frac{2}{9}$.